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0

After looking for an answer on another website, this formula is right.

2

Here is a general proof for all parameters in an open domain. $$dr = adt+bdW:=r\big(k(\theta-x)+\frac12\sigma^2\big)dt+\sigma rdW.$$ Let $$u(r(s),s):=e^{-\int_t^sr}B(r(s),s,T)=:\phi(s) B.$$ Then $$u(r(t),t)=\mathbf E\big[u(r(s),s)\big|r(t)\big],\, \forall t<s. \tag{1}$$ So, by Ito's Lemma, \begin{align} du(r(s),s) &= Bd\phi +\phi dB \\ &= \phi ...

1

We shall prove this by contradiction. Let $\theta=0$ and $\sigma=0$. $X_t=X_0e^{-kt}$ and $$B(0,t)=\exp\Big(-\int_0^te^{X_0e^{-ks}}ds\Big).$$ Suppose the contrary that $B(0,t)$ is affine. We should have $$B(0,t)=\exp{\left(A(0,t)-C(0,t)e^{X_0}\right)}\;\;\ \forall (t,X_0), \tag{1}$$ Differentiate the logarithm of Equation (1) with respect to $t$ side, ...

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